# Find the integrating factor of IDE,

First of all, the question is not about the general solution. My question on my book asks me to find the integrating factor.

Suppose I have simple IDE

$$(\sin y)\mathbb dx +(\cos x)\mathbb dy=0$$

I don't know how to find the integrating factor $$(\mu)$$ manually.

What i've done so far is :

Suppose it has form

$$M(x,y) \mathbb dx+N(x,y) \mathbb dy=0$$

Case 1 : $$\mu$$ is a function of $$x$$ only

\begin{aligned} \mu&=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\\ &=\frac{\cos y +\sin x}{\cos x} \end{aligned}

It doesn't work, cz still remaining the $$y$$

Case 2 : $$\mu$$ is a function of $$y$$ only

\begin{aligned} \mu&=\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\\ &=\frac{-\sin x -\cos y}{\sin y} \end{aligned}

Still remaining the $$x$$

Case 3 : $$\mu$$ is a function of $$xy$$

\begin{aligned} \mu&=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{My - Nx}\\ &=\frac{\cos y +\sin x}{y \sin x-x \sin y} \end{aligned}

I have no idea for this one...

But, when i use trial and error, i found

$$\mu=\csc y\cdot \sec x$$

But it's just a guess.

How to find the integrating factor systematically? I mean using such $$\mu=e^{\int f(x)\,\mathbb dx}$$?

Help me, and thanks.

• The exponential integral version of the integrating factor, I believe, only works for linear DE's, which this is not. – Adrian Keister Sep 27 '19 at 13:27
• Actually i found on youtube that is could be... but i don't know what the reference is. – user516076 Sep 27 '19 at 13:30
• There is no general recipe. If you assume $\mu=\mu(x)$, then you get $M_y \mu = N_x \mu + N \mu'$ so that $\ln(\mu)'=\frac{M_y - N_x}{N}$ which only helps if that is just a function of $x$. There's an analogous way to try $\mu$ being a function of just $y$. But if it is a function of both then you have $M_y \mu + M \mu_y = N_x \mu + N \mu_x$ which is too general to be useful without some further idea on what $\mu$ might look like. – Ian Sep 27 '19 at 13:36
• That said, a situation like this one where an integrating factor makes it separable is one to keep in mind; that one amounts to requiring $(M\mu)_y=(N\mu)_x=0$. – Ian Sep 27 '19 at 13:40
• You can view it as a pure eyeball guess, or you can view it slightly more systematically as "can I make this separable?" and then try to solve $(M\mu)_y=(N\mu)_x=0$. (This particular problem also fits into a general framework of $M(y) dx + N(x) dy = 0 \Rightarrow \mu=(MN)^{-1}$.) – Ian Sep 27 '19 at 13:41

$$\sin y \,dx + \cos x dy =0 ~~~(1)\implies \frac{dx}{\cos x}+ \frac{dy}{\sin y}=0~~~(2).$$ Integrating, we get $$\ln|\tan(x)+\sec(x)|- \ln|\cot(y)+\csc(y)|=C.$$
Note it is as though $$IF = \dfrac{1}{\cos x \sin y}= \sec x \csc y,$$ which upon multiplying the ODE (1) makes it EXACT as (2) where $$\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}.$$
• @user516076 yes, it is simple ODE which is easily separable in $x$ and $y$. – Dr Zafar Ahmed DSc Sep 27 '19 at 13:27
• @ user516075interesting you may say that $IF=\frac{1}{\cos x \sin y}$ is the integrating facor which when multiplied to the ODE makes it EXACT as $\frac{\partial M }{\partial y} = \frac{\partial N}{\partial x}.$ So have got the correct IF – Dr Zafar Ahmed DSc Sep 27 '19 at 13:33